Question

A 9.106 gram sample of an organic compound containing C, H and O is analyzed by combustion analysis and 18.19 grams of CO₂ and 7.449 grams of H₂O are produced.

In a separate experiment, the molar mass is found to be 88.11 g/mol. Determine the empirical formula and the molecular formula of the organic compound.
What is the empirical formula?

Answer 1

no of mole of CO2 produced = w/mwt = 18.19/44 = 0.413 mole

1 mole CO2 = 1 mole C

no of mole of C in sample = 0.413 mole

no of mole of H2O produced = (7.449/18) = 0.414 mole

    1 mole H2O = 2 mole H

no of mole of H in sample = 0.414*2 = 0.828 mole

mass of O in sample = 9.106 - (0.413*12+0.828*1)

                     = 3.322 g

no of mole of O in sample = 3.322/16 = 0.208 mole

c = 0.413   , H = 0.828 , O = 0.208

simplest ratio

C = 0.413/0.208 = 2 , H = 0.828/0.208 = 4   , O = 0.208/0.208 = 1

EMPIRICAL FORMULA = C2H4O

empirical formula mass = 12*2+1*4+1*16 = 44

n = molarmass/empirical formula mass

   = 88.11/44

   = 2

Molecular formula = 2*(C2H4O)

                   = C4H8O2